c++ - 如何将原始指针包装到 shared_ptr 中并防止 shared_ptr 删除对象？

Question

我需要将原始指针包装到 shared_ptr 中，以便将其传递给函数。该函数在返回后不保留对输入对象的任何引用。

{
  MyClass i;
  shared_ptr<MyClass> p(&i);
  f(p);
  // BAD: shared_ptr will delete i.
}

如何防止shared_ptr删除引用的对象？

Answer 1

作为chris在评论中提到，写一个空的删除器:

#include <type_traits>

template <typename T>
struct empty_delete
{
    empty_delete() /* noexcept */
    {
    }

    template <typename U>
    empty_delete(const empty_delete<U>&,
        typename std::enable_if<
            std::is_convertible<U*, T*>::value
        >::type* = nullptr) /* noexcept */
    {
    }

    void operator()(T* const) const /* noexcept */
    {
        // do nothing
    }
};

使用示例:

#include <iostream>
#include <memory>

struct noisy
{
    noisy() { std::cout << "alive" << std::endl; }
    ~noisy() { std::cout << "dead" << std::endl; }

    noisy(const noisy&);
    noisy& operator=(const noisy&);
};

template <typename T>
void take(T& yours)
{
    std::cout << "Taking..." << std::endl;
    {
        auto mine = std::move(yours);
    }
    std::cout << "Took." << std::endl;
}

int main()
{
    std::unique_ptr<noisy> a(new noisy());
    std::shared_ptr<noisy> b(new noisy());
    std::unique_ptr<noisy, empty_delete<noisy>> c(new noisy());
    std::shared_ptr<noisy> d(new noisy(), empty_delete<noisy>());

    take(a);
    take(b);
    take(c);
    take(d);
}

输出:

alive
alive
alive
alive
Taking...
dead
Took.
Taking...
dead
Took.
Taking...
Took.
Taking...
Took.

当然，这个例子会泄漏内存。